Question

Specifications for a critical part for a DVD player state that the part should weigh between 22 and 26 ounces. The process that produces the parts yields a mean of 24.5 ounces and a standard deviation of 0.85 ounce. The distribution of the weights of the part is normal.

a) What percentage of parts will meet the weight specifications?

b) If each production month 1,000,000 units of this part are produced at a cost of $50.00 per unit, what is the expected cost of the discarded units?

c) The DVD parts have a service life warranty of three years. The average life of the part is six years with.

d) Using the information in part C, give an estimate of the replacement cost due to product failure within the warranty period, if 1,000,000 units are sold. Assume that the cost per unit to produce the part is $50.

Hint: this is an application of the normal distribution.

Can this be answered in an excel sheet?

Answer 1

a)

Lower Specification Limit, LSL = 22 oz.

Upper Specification Limit, USL = 26 oz.

Process mean, m = 24.5 oz.

Std dev, s = 0.85 oz.

For LSL, z value = (22-24.5)/0.85 = -2.941, P(z) = NORMSDIST(-2.941) = 0.0016

For USL, z value = (26-24.5)/0.85 = 1.765, P(z) = NORMSDIST(1.765) = 0.9612

Percentage of parts that will meet the weight specifications = 0.9612 - 0.0016 = 0.9596 = 95.96 %

b) Expected cost of discarded units = (1-0.9596)*1000000*50 = $ 2,020,000

c) or part c and d, we need the standard deviation of life span.

Excel model is following:

LSL = USL m 26 24.5 0.85 6 P(Z1)0.0016 P(Z20.9612 z1--2.941 z2 1.765 10a) Percentage of parts that will meet the weight specifications- 95.96% 12 b) Expected cost of discarded units2,020,000 $ 13 14 c) 15 Average life 16 Standard deviation- 17 Warranty period- 18 19 20 21 Fraction of parts that will fail within warranty period years years years 6 standard deviation is not given in the question, hypothetical value is considered to show calculation P(z)0.1587 0.1587 23 d) Estimated replacement cost7,935,000 $ 25

EXCEL FORMULAS:

LSL =	22
USL =	26
m =	24.5
s =	0.85
z1 =	-2.941	P(z1) =	0.0016
z2 =	1.765	P(z2) =	0.9612
a) Percentage of parts that will meet the weight specifications =						95.96%
b) Expected cost of discarded units =			2,020,000	$
c)
Average life =		6	years
Standard deviation =		3	years	standard deviation is not given in the question, hypothetical value is considered to show calculation
Warranty period =		3	years
z =	-1	P(z) =	0.1587
Fraction of parts that will fail within warranty period =					0.1587
d) Estimated replacement cost =			7,935,000	$

Solution of part c and d using standard deviation of 1.0 year is following:

LSL = USL = 26 m 24.5 0.85 6 P(Z1)0.0016 P(Z2)0.9612 z1--2.941 Z2- 1.765 10 a) Percentage of parts that will meet the weight specifications- 95.96% 12 b) Expected cost of discarded units = 2,020,000 $ 13 14 c) 15 Average life= 16 Standard deviation- 17 Warranty period - 18 19 20 21 Fraction of parts that will fail within warranty period-0.0013 years year years 6 P(z)0.0013 23 d) Estimated replacement cost- 24 25 65,000 $